Given $$ z=x \cos y-y \cos x $$ find $$ \frac{d z}{d x}, \frac{d z}{d y}, \frac{d^{2} z}{d x^{2}}, \frac{d^{2} z}{d y^{2}}, \frac{d^{2} z}{d y d x} $$, and $$ \frac{d^{2} z}{d x d y} $$

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Alright, I need to find the first and second partial derivatives of the function $$ z = x \cos y - y \cos x $$ with respect to $$ x $$ and $$ y $$. Let's start by recalling the definitions and applying the necessary differentiation rules.

First, let's find $$ \frac{dz}{dx} $$. To do this, I'll differentiate $$ z $$ with respect to $$ x $$, treating $$ y $$ as a constant.

Given:
$$ z = x \cos y - y \cos x $$

Differentiating term by term:
1. The derivative of $$ x \cos y $$ with respect to $$ x $$ is $$ \cos y $$ because $$ \cos y $$ is treated as a constant.
2. The derivative of $$ -y \cos x $$ with respect to $$ x $$ is $$ y \sin x $$ because $$ y $$ is treated as a constant and the derivative of $$ \cos x $$ is $$ -\sin x $$.

So,
$$ \frac{dz}{dx} = \cos y + y \sin x $$

Next, let's find $$ \frac{dz}{dy} $$. This time, I'll differentiate $$ z $$ with respect to $$ y $$, treating $$ x $$ as a constant.

Given:
$$ z = x \cos y - y \cos x $$

Differentiating term by term:
1. The derivative of $$ x \cos y $$ with respect to $$ y $$ is $$ -x \sin y $$ because the derivative of $$ \cos y $$ is $$ -\sin y $$.
2. The derivative of $$ -y \cos x $$ with respect to $$ y $$ is $$ -\cos x $$ because $$ \cos x $$ is treated as a constant.

So,
$$ \frac{dz}{dy} = -x \sin y - \cos x $$

Now, let's find the second partial derivatives.

First, $$ \frac{d^2 z}{dx^2} $$:
Differentiate $$ \frac{dz}{dx} = \cos y + y \sin x $$ with respect to $$ x $$:
$$ \frac{d^2 z}{dx^2} = \frac{d}{dx}(\cos y) + \frac{d}{dx}(y \sin x) $$
Since $$ \cos y $$ is treated as a constant with respect to $$ x $$, its derivative is 0.
The derivative of $$ y \sin x $$ with respect to $$ x $$ is $$ y \cos x $$.

So,
$$ \frac{d^2 z}{dx^2} = 0 + y \cos x = y \cos x $$

Next, $$ \frac{d^2 z}{dy^2} $$:
Differentiate $$ \frac{dz}{dy} = -x \sin y - \cos x $$ with respect to $$ y $$:
$$ \frac{d^2 z}{dy^2} = \frac{d}{dy}(-x \sin y) + \frac{d}{dy}(-\cos x) $$
The derivative of $$ -x \sin y $$ with respect to $$ y $$ is $$ -x \cos y $$.
Since $$ \cos x $$ is treated as a constant with respect to $$ y $$, its derivative is 0.

So,
$$ \frac{d^2 z}{dy^2} = -x \cos y + 0 = -x \cos y $$

Now, let's find $$ \frac{d^2 z}{dy dx} $$ and $$ \frac{d^2 z}{dx dy} $$. These are the mixed partial derivatives, and for functions with continuous second derivatives, they should be equal.

First, $$ \frac{d^2 z}{dy dx} $$:
Differentiate $$ \frac{dz}{dy} = -x \sin y - \cos x $$ with respect to $$ x $$:
$$ \frac{d^2 z}{dy dx} = \frac{d}{dx}(-x \sin y) + \frac{d}{dx}(-\cos x) $$
The derivative of $$ -x \sin y $$ with respect to $$ x $$ is $$ -\sin y $$.
The derivative of $$ -\cos x $$ with respect to $$ x $$ is $$ \sin x $$.

So,
$$ \frac{d^2 z}{dy dx} = -\sin y + \sin x $$

Similarly, $$ \frac{d^2 z}{dx dy} $$:
Differentiate $$ \frac{dz}{dx} = \cos y + y \sin x $$ with respect to $$ y $$:
$$ \frac{d^2 z}{dx dy} = \frac{d}{dy}(\cos y) + \frac{d}{dy}(y \sin x) $$
The derivative of $$ \cos y $$ with respect to $$ y $$ is $$ -\sin y $$.
The derivative of $$ y \sin x $$ with respect to $$ y $$ is $$ \sin x $$.

So,
$$ \frac{d^2 z}{dx dy} = -\sin y + \sin x $$

Therefore, both mixed partial derivatives are equal:
$$ \frac{d^2 z}{dy dx} = \frac{d^2 z}{dx dy} = -\sin y + \sin x $$

To summarize, the derivatives are:
1. $$ \frac{dz}{dx} = \cos y + y \sin x $$
2. $$ \frac{dz}{dy} = -x \sin y - \cos x $$
3. $$ \frac{d^2 z}{dx^2} = y \cos x $$
4. $$ \frac{d^2 z}{dy^2} = -x \cos y $$
5. $$ \frac{d^2 z}{dy dx} = -\sin y + \sin x $$
6. $$ \frac{d^2 z}{dx dy} = -\sin y + \sin x $$

$$ \frac{dz}{dx} = \cos y + y \sin x $$ $$ \frac{dz}{dy} = -x \sin y - \cos x $$ $$ \frac{d^2 z}{dx^2} = y \cos x $$ $$ \frac{d^2 z}{dy^2} = -x \cos y $$ $$ \frac{d^2 z}{dy dx} = -\sin y + \sin x $$ $$ \frac{d^2 z}{dx dy} = -\sin y + \sin x $$ \]

Given \( z=x \cos y-y \cos x \) find \( \frac{d z}{d x}, \frac{d z}{d y}, \frac{d^{2} z}{d x^{2}}, \frac{d^{2} z}{d y^{2}}, \frac{d^{2} z}{d y d x} \), and \( \frac{d^{2} z}{d x d y} \)

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